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Homework Statement
Let U and V be open sets in Rn and let f be a one-to-one mapping from U onto V (so that there is an inverse mapping f-1). Suppose that f and f-1 are both continuous. Show that for any set S whose closure is contained in U we have f(bd(S)) = bd(f(S)).
Homework Equations
Open sets: every point in the set is an interior point. int(S) = S. Or S contains none of its boundry points.
bd(S) = {x in Rn | B(r,x)∩S≠ø and B(r,x)∩Sc≠ø for every r>0}
The Attempt at a Solution
ie, show that the function takes a boundry to a boundry.
Let S be a set in U. The let f(S) = T. Note that f-1(T) = S.
Also, since T = {x in Rn s.t. f(x) V} and f conts, and V open, then T is open.
The same conclusion holds for S.
Thus T and S are both open. They also have the same number of elements.
I don't know where to go from here.